How to understand this vector calculus idenity. Hello I am wondering how  to understand and if there is a not to advanced proof of the following:
If a vector $F$ is defined by three scalar functions $a$, $b$ and $c$ by $F=c(\nabla a \times \nabla b)$ then $div(F)=0$ provided that c is a function of a and b.
I know that if c is a function of a and b then they will have a zero jacobian but I don't know how else I can use that
Any help?
 A: Using the abbreviation $\partial_i = \partial/\partial x_i$ we have 
$$
\DeclareMathOperator{div}{div}
\begin{align}
\div F
&= \partial_i F_i \\ 
&= 
\partial_i(c(\nabla a \times \nabla b)_i) \\
&= 
(\partial_i c)(\nabla a \times \nabla b)_i + 
c (\partial_i)(\nabla a \times \nabla b)_i \\
&= \det(\nabla c, \nabla a, \nabla b) + c \det(\nabla, \nabla a, \nabla b) \\
&= A + B
\end{align}
$$
where we applied the product rule. The first term
$$
A = \det(\nabla c, \nabla a, \nabla b) 
$$
would vanish, if $\nabla a, \nabla b, \nabla c$ are linear dependent.
Using the total differential, we have
$$
\partial_i c = (\partial_a c) \, (\partial_i a) + (\partial_b c) \, (\partial_i b) \iff \\
\nabla c = (\partial_a c) \nabla a + (\partial_b c) \nabla b
$$
so indeed those three gradient vectors are linear dependent.
For the second term I have only a complicated argument:
Component-wise we get:
$$
\begin{align}
B & = c \det(\nabla, \nabla a, \nabla b) \\
&= c (\partial_i)(\nabla a \times \nabla b)_i \\
&= c (\partial_i)(\epsilon_{ijk} (\partial_j a)(\partial_k b)) \\
&= c \, \epsilon_{ijk} 
((\partial_i\partial_j a) (\partial_k b) + 
 (\partial_j a) (\partial_i\partial_k b)) \\
\end{align}
$$
where we used $(a \times b)_i = \epsilon_{ijk} a_j b_k$ together with the Einstein summation convention to sum over indices which show up twice and the product rule $(fg)' = f'g + f g'$ being applied. Then
$$
\begin{align}
B
&= c \, (\epsilon_{ijk} 
(\partial_i\partial_j a) (\partial_k b)) + 
c(\epsilon_{ijk}(\partial_j a) (\partial_i\partial_k b)) \\
&= c \, ((\partial_k b) \epsilon_{kij} 
(\partial_i\partial_j a) ) + 
c((\partial_j a)\epsilon_{jki}(\partial_i\partial_k b)) \\
&= c \, ((\partial_k b) \epsilon_{kij} 
(\partial_i\partial_j a) ) - 
c((\partial_j a)\epsilon_{jik}(\partial_i\partial_k b)) \quad (*) \\
&= c \det(\nabla b, \nabla, \nabla a) - c \det(\nabla a, \nabla, \nabla b)  \\ 
\end{align}
$$
Here we did a little reordering and used that we can perform a cyclic shift on a permutation, without changing its sign: $\epsilon_{ijk} = \epsilon_{jki}$ plus for the second sum that if we transpose two indices then we flip the sign of the permutation: $\epsilon_{ijk} = - \epsilon_{ikj}$.
We conclude
$$
\begin{align}
B 
&= 
c \, ((\partial_k b) \epsilon_{kij} (\partial_j\partial_i a) ) - 
c((\partial_j a)\epsilon_{jik}(\partial_k\partial_i b)) \\
\end{align}
$$
because $\partial_i \partial_j = \partial_j \partial_i$ (see here). Again transposing the permutations gives
$$
\begin{align}
B 
&= 
c \, ((\partial_k b) (-\epsilon_{kji}) (\partial_j\partial_i a) ) - 
c((\partial_j a)(-\epsilon_{jki})(\partial_k\partial_i b)) \\
&= 
c \, ((\partial_k b) (-\epsilon_{kij}) (\partial_i\partial_j a) ) - 
c((\partial_j a)(-\epsilon_{jik})(\partial_i\partial_k b)) \quad (**) \\
&= -c \det(\nabla b, \nabla, \nabla a) + c \det(\nabla a, \nabla, \nabla b)  
\end{align}
$$
where for the last step we renamed the indices such that $i \leftrightarrow j$ for the first sum and $i\leftrightarrow k$ for the second sum.
But comparing $(*)$ with $(**)$ we note that the sums behave like $s = -s \iff 2s = 0 \iff s = 0$. So both sums vanish and $B$ vanishes too.
A: Since $c$ is a function of $a$ and $b$, the Chain Rule says
$$
\nabla c=\frac{\partial c}{\partial a}\nabla a+\frac{\partial c}{\partial b}\nabla b\tag{1}
$$
Therefore,
$$
\begin{align}
\nabla c\cdot(\nabla a\times\nabla b)
&=\left(\frac{\partial c}{\partial a}\nabla a+\frac{\partial c}{\partial b}\nabla b\right)\cdot(\nabla a\times\nabla b)\\
&=\frac{\partial c}{\partial a}\nabla a\cdot(\nabla a\times\nabla b)
+\frac{\partial c}{\partial b}\nabla b\cdot(\nabla a\times\nabla b)\\[6pt]
&=0+0\tag{2}
\end{align}
$$
Furthermore,
$$
\begin{align}
&\nabla\cdot(\nabla a\times\nabla b)\\[6pt]
&=
\frac{\partial}{\partial x_1}\left(\frac{\partial a}{\partial x_2}\frac{\partial b}{\partial x_3}-\frac{\partial a}{\partial x_3}\frac{\partial b}{\partial x_2}\right)\\
&+
\frac{\partial}{\partial x_2}\left(\frac{\partial a}{\partial x_3}\frac{\partial b}{\partial x_1}-\frac{\partial a}{\partial x_1}\frac{\partial b}{\partial x_3}\right)\\
&+
\frac{\partial}{\partial x_3}\left(\frac{\partial a}{\partial x_1}\frac{\partial b}{\partial x_2}-\frac{\partial a}{\partial x_2}\frac{\partial b}{\partial x_1}\right)\\[6pt]
&=0\tag{3}
\end{align}
$$
because the coefficients of $\frac{\partial a}{\partial x_k}$ cancel and the coefficients of $\frac{\partial b}{\partial x_k}$ cancel. For example, the coefficient of $\frac{\partial a}{\partial x_1}$ is $\frac{\partial^2b}{\partial x_3\partial x_2}-\frac{\partial^2b}{\partial x_2\partial x_3}=0$ and the coefficient of $\frac{\partial b}{\partial x_1}$ is$\frac{\partial^2a}{\partial x_2\partial x_3}-\frac{\partial^2a}{\partial x_3\partial x_2}=0$.
Therefore,
$$
\begin{align}
\nabla\cdot(c(\nabla a\times\nabla b))
&=\nabla c\cdot(\nabla a\times\nabla b)+c\nabla\cdot(\nabla a\times\nabla b)\\[6pt]
&=0\tag{4}
\end{align}
$$
